Step-by-step explanation:
The solution to this problem is very much similar to your previous ones, already answered by Sqdancefan.
Given:
mean, mu = 3550 lbs (hope I read the first five correctly, and it's not a six)
standard deviation, sigma = 870 lbs
weights are normally distributed, and assume large samples.
Probability to be estimated between W1=2800 and W2=4500 lbs.
Solution:
We calculate Z-scores for each of the limits in order to estimate probabilities from tables.
For W1 (lower limit),
Z1=(W1-mu)/sigma = (2800 - 3550)/870 = -.862069
From tables, P(Z<Z1) = 0.194325
For W2 (upper limit):
Z2=(W2-mu)/sigma = (4500-3550)/879 = 1.091954
From tables, P(Z<Z2) = 0.862573
Therefore probability that weight is between W1 and W2 is
P( W1 < W < W2 )
= P(Z1 < Z < Z2)
= P(Z<Z2) - P(Z<Z1)
= 0.862573 - 0.194325
= 0.668248
= 0.67 (to the hundredth)
Exact form:
22/13
Mixed number:
1 9/13
D is the answer hope this helps
Answer:
Step-by-step explanation:
There are no categorical antonyms for eleven. The numeral eleven is defined as: The cardinal number occurring after ten and before twelve.
Answer:
Area of ΔDEF is 
.
Step-by-step explanation:
Given;
ΔABC and ΔDEF is similar and ∠B ≅ ∠E.
Length of AB = 
and
Length of DE = 
Area of ΔABC = 
Solution,
Since, ΔABC and ΔDEF is similar and ∠B ≅ ∠E.
Therefore,
Where triangle 1 and triangle 2 is ΔABC and ΔDEF respectively.
If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.
Thus the area of ΔDEF is .
